what happens when chloroform is boiled with aqueous solution of custic potash
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When chloroform reacts with aq. KOH, the chlorines on the carbon atom are successively replaced by -OH groups from KOH via. nucleophilic reaction (SN2). In theory, first it forms CHCl2(OH), then CHCl(OH2), and then CH(OH)3, while eliminating KCl with each step. Since Cl- is an excellent leaving group, CHCl(OH2) will spontaneously release one molecule of H2O and produce HC(=O)Cl or formyl chloride, which again will be hydrolyzed to produce HCOOK or Potassium Formate.
1 mole of Chloroform will react with 4 moles of KOH to produce 1 mole of HCOOK (potassium formate), 3 moles of KCl and 2 moles of H2O.
1 CHCl3 + 4 KOH —-> HCOOK + 3 KCl + 2 H2O.
This is if you allow the reaction to go to completion. If the reaction conditions are not severe enough, you can isolate some of the intermediates as well.
HOPE IT HELPS YOU AND MARK AS BRAINLIEST.
When chloroform reacts with aq. KOH, the chlorines on the carbon atom are successively replaced by -OH groups from KOH via. nucleophilic reaction (SN2). In theory, first it forms CHCl2(OH), then CHCl(OH2), and then CH(OH)3, while eliminating KCl with each step. Since Cl- is an excellent leaving group, CHCl(OH2) will spontaneously release one molecule of H2O and produce HC(=O)Cl or formyl chloride, which again will be hydrolyzed to produce HCOOK or Potassium Formate.
1 mole of Chloroform will react with 4 moles of KOH to produce 1 mole of HCOOK (potassium formate), 3 moles of KCl and 2 moles of H2O.
1 CHCl3 + 4 KOH —-> HCOOK + 3 KCl + 2 H2O.
This is if you allow the reaction to go to completion. If the reaction conditions are not severe enough, you can isolate some of the intermediates as well.
HOPE IT HELPS YOU AND MARK AS BRAINLIEST.
Answered by
17
Hi
When chloroform reacts with aq. KOH, the chlorines on the carbon atom are successively replaced by -OH groups from KOH via. nucleophilic reaction (SN2). In theory, first it forms CHCl2(OH), then CHCl(OH2), and then CH(OH)3, while eliminating KCl with each step. Since Cl- is an excellent leaving group, CHCl(OH2) will spontaneously release one molecule of H2O and produce HC(=O)Cl or formyl chloride, which again will be hydrolyzed to produce HCOOK or Potassium Formate.
CHCl3 + 4 KOH —-> HCOOK + 3 KCl + 2 H2O.
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